Examples

  • An application with mass acceleration factor 2.5 (load classification II), 14 hours daily operating time ( read at 16 h/day) and 300 cycles/h results in the service factor fB_L = 1.5 according to the diagram. The fB value of the required gearmotor must therefore be ≥ 1.5.
    • If the gearmotor is intended for operation at -35 °C, the following applies:
    • fB_req = fB_L × fB3 = 1.5 × 1.2 = 1.8
    • The gearmotor to be selected now requires an fB value of ≥ 1.8.
  • The gearmotor with service factor fB_L = 1.5 of the previous example is to be a helical-worm gearmotor, and the ambient temperature is 40 °C:
    • fB1 = 1.36 (read at load classification II)
    • Time under load = 40 min/h → cdf = 66.67% → fB2 = 0.95
    • The required service factor is:
    • fB_req = fB_L × fB1 × fB2 = 1.5 × 1.36 × 0.95 = 1.94
    • The selected helical-worm gearmotor requires a service factor fB ≥ 1.94.